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Let $N = \{1,(12)(34),(13)(24),(14)(23)\}$. Determine if the quotient group $S_4/N$ is isomophic to $S_3$.

I computed the cosets: $N, (12)N, (13)N, (14)N, (123)N, (234)N$, and the others are equivalent to one of these. This took me a while to do. Is there a faster way and what do I do next?

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It suffices you show $S_4/N$ is nonabelian, since $S_3$ is the only nonabelian group of order $6$ (which is $24/4=S_4/V_4$). Now $S_4/N$ abelian implies that $N$ contains the commutator subgroup of $S_4$; but this is $A_4$, which has order $12$. Since $12>4$, this cannot be, and $S_4/N$ is nonabelian.

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  • $\begingroup$ How do I computer S_4 /N quickly? $\endgroup$ – Lin Feb 4 '16 at 23:02
  • $\begingroup$ @Lin By using a computer :) $\endgroup$ – user26857 Feb 4 '16 at 23:30
  • $\begingroup$ So I have to manually compute every coset... Oh man I wish algebra was nice to me $\endgroup$ – Lin Feb 4 '16 at 23:51

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